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I am reviewing properties of atoms and I find myself becoming uncomfortable with spin orbit coupling. For ease, let's consider the case of hydrogen. In particular, at very small $r$,

$$H_{\mathrm{SO}} = \frac{c}{r^3} \vec{S}\cdot \vec{L} \gg \frac{\hbar^2 l(l+1)}{2m r^2}=H_{\mathrm{Cent}}$$

This seems to me to be a problem when $\vec{S}\cdot{\vec{L}}$ is negative [i.e. when $j(j+1)<l(l+1)+3/4$, so when the spin and angular momentum are anti-aligned], as it means that attraction from spin-orbit coupling can overcome the centripetal barrier. For example, I would naively expect that one could variationally show that the energy is unbounded below. This leads me to the following question:

Is the lowest energy eigenvalue of the naive Hamiltonian for the hydrogen atom, including spin-orbit coupling as the only perturbation to the coulomb potential, in fact unbounded below?

If this is the case, I will ask a separate question about the resolution to this problem. I suspect the inclusion of perturbative terms of relativistic origin that will further penalize wavefunctions that are sharply peaked will be important. However, I would prefer to keep these considerations out of this problem and instead to focus on whether taking spin-orbit coupling in the form $H_{\mathrm{SO}} = \frac{c}{r^3} \vec{S}\cdot \vec{L}$ as the only additional term to the potential will cause the lowest energy to be unbounded below.

Buzz
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user196574
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2 Answers2

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As you say, the spin-orbit interaction is a perturbation. This means that the perturbation Hamiltonian $H_{SO} = \vec{S} \cdot \vec{L} / r^3$ is such that the corresponding energy corrections are much smaller than the unpeturbed energy levels (i.e. the Bohr energy levels). This seems to be case for the Hydrogen atom, as effectively the electron is on average at a Bohr radius from the proton. It is also likely that the expression for $H_{SO}$ should not hold all the way to $r \to 0$, as the proton is not a point-like particle.

Miguel Correia
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  • Thank you for the response. I think that indeed a modified $H_{SO}$ that includes e.g. relativistic effects or takes into account the radius of the proton could truly be a small perturbation. However, my question is if we add in the naive spin-orbit coupling as a term in the Hamiltonian, will the Hamiltonian be bounded below? Your answer doesn't yet answer this question. – user196574 Apr 14 '22 at 02:23
  • Yes in that case you can have metastable states at large radius, as the 1/r^3 term becomes subleading. However, as you say, at small radius that term will dominate and can make the potential unbounded, meaning that any metastable state will eventually tunnel to r->0 and decay. So yes there wont be a real spectrum of energies in that case. – Miguel Correia Apr 14 '22 at 11:12
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(Edited after comments from the OP.)

$H_{so}$ is indeed unbounded below (for $\ell > 0$ states), but the matrix elements you need for perturbative calculations are all finite and not sensitive to the exact behavior of $H_{so}$ near the origin.

So, as Miguel Correia suggests, it makes physical sense to "cut off" $H_{so}$ near the origin. This would make it bounded but not affect the individual matrix elements much.

To see that the $-1/r^3$ Hamiltonian is unbounded below, use the $n=0$, $\ell=1$ harmonic oscillator as a trial wavefunction: $R_{01} = \frac{2^{3/2}}{3^{1/2}} \frac{1}{\pi^{1/4}} \beta^{5/2} r \exp(-\beta^2r^2/2)$. The expectation value of $-1/r^3$ is then $-8\beta^3/3\sqrt{\pi}$, which can be made arbitrarily negative by choosing beta large. (And the kinetic energy term is only proportional to $\beta^2$, so it can't compensate.)

Regarding the matrix elements: $H_{so}$ between two s-wave states is zero by parity, so just consider matrix elements where at least one of the states has $\ell > 0$. Recall that the wavefunctions for angular momentum $\ell$ are proportional to $r^\ell$ near the origin, so the integrand in the matrix element has a factor of $r^{\ell_1} r^{\ell_2} r^2$ (where the $r^2$ comes from the volume element $r^2 dr$), which kills the $r^{-3}$ in $H_{so}$.

Paul G
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  • Just because the diagonal and off-diagonal matrix elements are finite does not mean that the Hamiltonian is bounded below. For example, the $m$ by $m$ matrix whose entries are all $-1$ has an eigenvalue of $-m$, which, in the limit as $m \rightarrow \infty$, is unbounded below. This is relevant because there would be an infinite number of nonzero entries in the Hydrogen atom + spin-orbit Hamiltonian, so the boundedness of individual entries is not enough to show that the Hamiltonian is bounded below. In fact, I still suspect it is not bounded below. – user196574 Apr 14 '22 at 21:27
  • Yes, you are right that the $-1/r^3$ Hamiltonian is unbounded below: Using the $n=0$, $\ell=1$ harmonic oscillator as a trial wavefunction, $R_{01} = \frac{2^{3/2}}{3^{1/2}} \frac{1}{\pi^{1/4}} \beta^{5/2} r \exp(-\beta^2r^2/2)$, the expectation value of $-1/r^3$ is $-8\beta^3/3\sqrt{\pi}$, which can be made arbitrarily negative by choosing a large beta. (And the positive kinetic energy term is only proportional to $\beta^2$, so it can't compensate.) – Paul G Apr 15 '22 at 01:04
  • Thanks, please add that to your answer and I'll accept it. – user196574 Apr 15 '22 at 01:30